provide solution for RD Sharma maths class 12 chapter Indefinite Integrals exercise  18.9 question 46

Answer: $\frac{1}{4} \sin ^{4} x-\sin ^{2} x+\log |\sin x|+c$

Hint: Use substitution method to solve this integral.

Given:   $\int \frac{\cos ^{5} x}{\sin x} d x$

Solution:

$\text { Let } I=\int \frac{\cos ^{5} x}{\sin x} d x$

\begin{aligned} &\text { Put } \sin x=t \Rightarrow \cos x \; d x=d t \\ &\Rightarrow d x=\frac{d t}{\cos x} \text { then } \end{aligned}

$I=\int \frac{\cos ^{5} x}{t} \frac{d t}{\cos x}=\int \frac{\cos ^{4} x}{t} d t$

$=\int \frac{\left(\cos ^{2} x\right)^{2}}{t} d t=\int \frac{\left(1-\sin ^{2} x\right)^{2}}{t} d t$                        $\left[\begin{array}{l} \because \cos ^{2} x+\sin ^{2} x=1 \\ \Rightarrow \cos ^{2} x=1-\sin ^{2} x \end{array}\right]$

$=\int \frac{\left(1-t^{2}\right)^{2}}{t} d t$                    $[\because \sin x=t]$

$=\int\left\{\frac{1+\left(t^{2}\right)^{2}-2 t^{2}}{t}\right\} d t$            $\left[\because(a-b)^{2}=a^{2}+b^{2}-2 a b\right]$

\begin{aligned} &=\int\left\{\frac{1+\left(t^{4}\right)-2 t^{2}}{t}\right\} d t \\ &=\int\left\{\frac{1}{t}+\frac{t^{4}}{t}-\frac{2 t^{2}}{t}\right\} d t \\ &=\int\left\{\frac{1}{t}+t^{3}-2 t\right\} d t \end{aligned}

$=\int \frac{1}{t} d t+\int t^{3} d t-2 \int t\; d t$

$=\log |t|+\frac{t^{3+1}}{3+1}-2 \frac{t^{1+1}}{1+1}+c$                        $\left[\because \int x^{n} d x=\frac{x^{n+1}}{n+1}+c\right]$

\begin{aligned} &=\log |t|+\frac{t^{4}}{4}-2 \frac{t^{2}}{2}+c \\ &=\frac{1}{4} \sin ^{4} x-\sin ^{2} x+\log |\sin x|+c \quad[\because t=\sin x] \end{aligned}